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Chapter 10: Problem 54

A vertical cylinder of crosssectional area \(A\) is fitted with \(\underline{a}\) tight-fitting, frictionless piston of mass \(m\) (Fig. P10.54). (a) If \(n\) moles of an ideal gas are in the cylinder at a temperature of \(T\), use Newton's second law for equilibrium to show that the height \(h\) at which the piston is in equilibrium under its own weight is given by $$ h=\frac{n R T}{m g+P_{n} A} $$ where \(P_{0}\) is atmospheric pressure. (b) Is the pressure inside the cylinder less than, equal to, or greater than atmospheric pressure? (c) If the gas in the cylinder is warmed, how would the answer for \(h\) be affected?

Short Answer

Expert verified
a) The height \(h\) of the piston at equilibrium is given by \(h=\frac{n R T}{P_{0} A+m g}\).\nb) The pressure inside the cylinder is less than the atmospheric pressure.\nc) If the gas is warmed, the height \(h\) would increase.

Step by step solution

01

Apply Newton's Second Law for equilibrium

Start by applying the second law of Newton to the piston. The total force exerted on the piston must balance the weight of the piston itself, i.e., the gravitational force acts downwards and the sum of atmospheric pressure and the pressure in the cylinder act upwards. This gives: \[ P_{n}A+m g=P_{0} A \] from which the pressure inside the cylinder can be isolated as follows: \[ P_{n}=P_{0}-\frac{m g}{A} \]
02

Use the Ideal Gas Law

Next, use the ideal gas law, which is given as \(P V=n R T\), to express the volume \(V\) in terms of \(n\), \(R\), \(T\), and \(P_{n}\). The volume is the area times the height \(h\), i.e., \(V=A h\). This can be substituted into the ideal gas law to obtain: \[ P_{n} A h=n R T \] From here, the height \(h\) can be isolated: \[ h=\frac{n R T}{P_{n} A} \]
03

Combine the Results

Equating equations in steps 1 and 2, we can obtain the required expression for \(h\) under equilibrium conditions. Substituting \(P_{n}=P_{0}-\frac{m g}{A}\) into \(h=\frac{n R T}{P_{n} A}\), we get: \[ h=\frac{n R T}{(P_{0}-\frac{m g}{A}) A}=\frac{n R T}{(P_{0} A+m g)} \]
04

Analyze the Pressure and Temperature Changes

(b) The pressure inside the cylinder is less than the atmospheric pressure, since we subtract the gravitational force due to the weight of the piston from \(P_{0}\).\n(c) If the gas in the cylinder is warmed, the temperature \(T\) increases. As \(T\) is in the numerator of the equation for \(h\), \(h\) would increase accordingly.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Newton's Second Law
Newton's Second Law is fundamental to understanding many physical phenomena, including the equilibrium conditions for forces acting on an object. In the context of the exercise, the piston in the cylinder is in equilibrium when the total forces acting on it are balanced. This involves the sum of the upward forces, due to both atmospheric pressure and the pressure from the gas inside the cylinder, being equal to the downward gravitational force exerted by the weight of the piston. This can be expressed as: \[ P_{n}A + mg = P_{0} A \] where:
  • \( P_{n} \) is the internal pressure of the gas.
  • \( A \) is the cross-sectional area of the cylinder.
  • \( mg \) is the gravitational force (the weight of the piston).
  • \( P_{0} \) is the atmospheric pressure.
By balancing these forces, we can solve for the internal pressure \( P_{n} \), understanding its value in terms of the external atmospheric pressure and the weight of the piston. This application of Newton's Second Law ensures the piston does not accelerate, maintaining a stable position.
Pressure Equilibrium
Pressure equilibrium is achieved when the forces due to pressure differences on either side of the piston are balanced, allowing the piston to stay stationary. In this exercise, the pressure inside the cylinder, exerted by the ideal gas, plays a vital role in maintaining this balance against external atmospheric pressure. The pressure inside the cylinder can be determined by rearranging the equilibrium condition: \[ P_{n} = P_{0} - \frac{mg}{A} \] Here, the pressure length \( P_{n} \) becomes less than the atmospheric pressure \( P_{0} \) due to the gravitational force from the piston's weight, as subtracted from \( P_{0} \).

This pressure equilibrium explains why the piston does not move upwards or downwards and provides insight into how any changes in variables like temperature can influence the position of the piston. If, for instance, more weight was added to the piston or if external pressure changed dramatically, this balance would be disrupted, affecting the height \( h \) of the piston.
Thermal Expansion of Gases
Thermal expansion of gases relates to how a gas expands when heated, causing variables like volume and pressure to adjust. According to the Ideal Gas Law \( PV = nRT \), an increase in temperature \( T \) leads to changes in volume and/or pressure to maintain the equation's equality. In the problem here, as the gas in the cylinder is warmed:
  • The temperature \( T \) increases.
  • The internal pressure \( P_{n} \) can initially remain the same.
  • The height \( h \), connected to volume \( V = Ah \), must increase to correspond with the rise in temperature if pressure is held constant.
This relation indicates that the piston's height \( h \) will rise as a result of increasing temperature. The increased thermal activity in the gas generates more force on the piston, pushing it upward and reflecting how gases expand when heated.

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Most popular questions from this chapter

The ideal gas law can be recast in terms of the density of a gas. (a) Use dimensional analysis to find an expression for the density \(\rho\) of a gas in terms of the number of moles \(n_{1}\) the volume \(V_{1}\) and the molecular weight \(M\) in kilograms per mole. (b) With the expression found in part (a), show that $$ P=\frac{\rho}{M} R T $$ for an ideal gas. (c) Find the density of the carbon dioxide atmosphere at the surface of Venus, where the pressure is \(90.0\) atm and the temperature is \(7.00 \times 10^{2} \mathrm{~K}\). (d) Would an evacuated steel shell of radius \(1.00 \mathrm{~m}\) and mass \(2.00 \times\) \(10^{2} \mathrm{~kg}\) rise or fall in such an atmosphere? Why?

Two small containers, each with a volume of \(100 \mathrm{~cm}^{3}\), contain helium gas at \(0^{-} \mathrm{C}\) and \(1.00\) atm pressure. The two containers are joined by a small open tube of negligible volume, allowing gas to flow from one container to the other. What common pressure will exist in the two containers if the temperature of one container is raised to \(100^{\circ} \mathrm{C}\) while the other container is kept at \(0^{\circ} \mathrm{C}\) ?

Show that the temperature \(-40^{\circ}\) is unique in that it has the same numerical value on the Celsius and Fahrenheit scales.

Death Valley holds the record for the highest recorded temperature in the United States. On July 10, 1913 , at at place called Fumace Creek Ranch, the temperature rose to \(134^{\circ} \mathrm{F}\). The lowest U.S. temperature ever recorded occurred at Prospect Creek Camp in Alaska on January 23,1971, when the temperature plummeted to \(-79.8^{\circ}\) F. (a) Convert these temperatures to the Celsius scale. (b) Convert the Cielsius temperatures to Kelvin.

A bimetallic bar is made of two thin strips of dissimilar metals bonded together. As they are heated, the one with the larger average coefficient of expansion expands more than the other, forcing the bar into an arc, with the outer strip having both a larger radius and a larger circumference (Fig. P10.61). (a) Derive an expression for the angle of bending, \(\theta\), as a function of the initial length of the strips. their average coefficientsof linear expansion, the change in temperature, and the separation of the centers of the strips \(\left(\Delta r=r_{2}-r_{1}\right) .\) (b) Show that the angle of bending goes to zero when \(\Delta T\) goes to zero or when the two coefficients of expansion become equal. (c) What happens if the bar is cooled?
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